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Inscribing Angles and Polygons in Circles


Inscribed Angle

An inscribed angle is the angle created when two chords or secants intersect a circle. The arc that lies between the two lines, rays, or segments is called an intercepted arc.

It is then said that the angle at AA intercepts the arc BC.BC.

Inscribed Angle Theorem

The measure of an inscribed angle is half that of its intercepted arc.

In the diagram above, the measure of BAC\angle BAC is half the measure of arc BC.BC. This can be proven with the Isosceles Triangle Theorem.

Given that the arc ADBADB measures 270,270 ^\circ, find the measure of the inscribed angles D\angle D and C.\angle C.

Show Solution

To find the angle measures, we can use the Inscribed Angle Theorem. Thus, we have to find the measure of arc ABAB first. The minor arc AB,AB, together with its corresponding major arc ADB,ADB, corresponds to a full rotation. Thus, the sum of their measures is 360.360^\circ. This means that the measure of ABAB must be 360270=90. 360^\circ - 270^\circ = 90^\circ. By the Inscribed Angles Theorem, angle D\angle D measures half that of its intercepted arc, AB.AB.

Thus, mD=902=45. m\angle D = \dfrac{90^\circ} 2 = 45^\circ. For the same reason, mCm\angle C is 4545 ^\circ as well.


Circumscribed Angle

When two tangents of a circle intersect to create an angle, it is called a circumscribed angle.

In the diagram above, AC\overleftrightarrow{AC} and AB\overleftrightarrow{AB} are tangents to the circle, and A\angle A is a circumscribed angle.

Polol is a newbie chopstick user. When he picks a certain sushi roll up, he notices that he crosses the chopsticks. Out of curiosity, he wants to know the circumscribed angle created by the crossed chopsticks. For this, he makes two measurements: the arc between the chopsticks' point of contact with the roll measures 4.74.7 cm and the sushi roll's diameter is 3.53.5 cm. Use his measurements to find the angle.

Show Solution

Let's start by drawing a diagram corresponding to the situation. Notice that the sushi roll can be modeled by a circle, with the chopsticks being tangents to the circle.

The diameter of the circle is then 3.53.5 cm, the length of arc BCBC is 4.74.7 cm, and we are interested in the measure of angle A.\angle A. To find this, we'll first need the measure of arc BC,BC, which we can find by the ratio between the arc length and the circumference. The circumference is πd=π3.511.0 cm. \pi d = \pi \cdot 3.5 \approx 11.0 \text{ cm.} We can now calculate the arc measure.

arc lengthcircumference=arc measure360\dfrac{\text{arc length}}{\text{circumference}} = \dfrac{\text{arc measure}}{360^\circ}
4.711.0=arc measure360\dfrac{{\color{#0000FF}{4.7}}}{{\color{#009600}{11.0}}} = \dfrac{\text{arc measure}}{360^\circ}
4.711.0360=arc measure\dfrac{4.7}{11.0} \cdot 360^\circ = \text{arc measure}
153.81818=arc measure153.81818\ldots^\circ = \text{arc measure}
154arc measure154^\circ \approx \text{arc measure}
arc measure154\text{arc measure} \approx 154^\circ

We now know that the measure of arc BCBC is 154.154^\circ. As ABOCABOC is a quadrilateral, it has the interior angle sum 360.360^\circ. With mBm\angle B and mCm\angle C being 90,90^\circ, the sum of mAm\angle A and mOm\angle O must then be 180.180^\circ.

mA+mO=180m\angle A + m\angle O = 180^\circ
mA+154=180m\angle A + 154^\circ = 180^\circ
mA=26m\angle A = 26^\circ

The measure of the circumscribed angle is 26.26^\circ.


Inscribed Polygon

A polygon whose vertices all lie on a circle is called an inscribed polygon. In this case, the circle is referred to as a circumscribed circle.


Inscribed Quadrilateral Theorem

The pairs of opposite angles in an inscribed quadrilateral are supplementary.

In the diagram above, mA+mC=180andmB+mD=180. m\angle A + m\angle C = 180^\circ \quad \text{and} \quad m\angle B + m\angle D = 180^\circ. This can be proven using the Inscribed Angle Theorem.


Inscribed Quadrilateral Theorem

Consider the circle and the inscribed quadrilateral ABCD.ABCD.

Notice that the arcs BCDBCD and BADBAD together span the entire circle. Together, the sum of their measures is 360.360^\circ. By the Inscribed Angle Theorem, the measure of arc BCDBCD is equal to twice the measure of A.\angle A. Similarly, the measure of BADBAD is twice that of mC.m \angle C. By substitution, this gives 2mC+2mA=360. 2m\angle C + 2m\angle A = 360^\circ. Simplifying gives mC+mA=180. m\angle C + m\angle A = 180^\circ.

Similar reasoning yields the same relationship between mBm\angle B and mD.m\angle D. Therefore, the pairs of opposite angles of an inscribed quadrilateral are supplementary.


Inscribing a Square in a Circle


Using a compass and straightedge, it's possible to construct a square that's inscribed in a given circle. Start by drawing any diameter, naming the endpoints AA and B.B.

Then, using the compass and straightedge, draw the perpendicular bisector of the diameter. Name the intersections between the bisector and the circle CC and D,D, respectively.

The quadrilateral ACBDACBD can now be drawn, which is in fact a square.



Drawing the Inscribed Circle of a Triangle


In a couple of steps, it's possible to draw the inscribed circle of a triangle, which is the largest circle completely contained within the triangle. First, draw the angle bisector of two vertices of the triangle. Label their intersection A.A.

This point, A,A, is called the incenter of the triangle and is the only point for which the shortest distance to each side is equal. Next, draw a line through AA that intersects one of the sides at a right angle. Label this intersection B.B.

Place the sharp end of the compass at AA and the pencil end at B.B. Draw a circle.

This circle is the inscribed circle of the triangle.


Drawing the Circumscribed Circle of a Triangle


Finding the circumscribed circle of a triangle, which is the circle intersecting all vertices, is done in a similar fashion as finding the inscribed one. Start by drawing the perpendicular bisector of two sides of the triangle. Label their intersection A.A.

This point, A,A, is called the circumcenter and is located at equal distance from the vertices of the triangle. Now, place the sharp end of the compass in A,A, the pencil end in any vertex, and draw the corresponding circle.

This circle is the circumscribed circle of the triangle.
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